1. Let ƒ : R → R where ƒ(x) = x², c = -1, and u = 2. Use the e-d definition to prove Lu=? (you need to find Lu in your scratchwork).
1. Let ƒ : R → R where ƒ(x) = x², c = -1, and u = 2. Use the e-d definition to prove Lu=? (you need to find Lu in your scratchwork).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:1. Let \( f : \mathbb{R} \rightarrow \mathbb{R} \) where \( f(x) = x^2 \), \( c = -1 \), and \( u = 2 \). Use the \( \epsilon \)-\( \delta \) definition to prove \( L_u = ? \) (you need to find \( L_u \) in your scratchwork).
![**Example:**
Let \( f: \mathbb{R}^2 \to \mathbb{R}^2 \) be defined as \( f(x, y) = \left( x^2 y, (3x + 2)y \right) \). Consider vectors \( u = (1, 0) \) and \( c = (1, 1) \).
**Claim:** \( Lu = (2, 3) \)
**Proof:**
Fix \(\epsilon > 0\). Choose \(\delta = \frac{\epsilon}{2}\). Assume \(0 < |t| < \delta\). Then:
\[
\frac{1}{t} \left[ f(c + tu) - f(c) - Lu \right] = \left \lVert \frac{1}{t} \left[ f(1+t, 1) - (1, 5) - (2, 3) \right] \right \rVert
\]
\[
= \left \lVert \frac{1}{t} \left[ ((1+t)^2, 3(1+t)+2) - (1, 5) \right] - (2, 3) \right \rVert
\]
\[
= \left \lVert \frac{1}{t} \left[ (1 + 2t + t^2, 3t + 5) - (1, 5) \right] - (2, 3) \right \rVert
\]
\[
= \left \lVert \frac{1}{t} \left[ (2t + t^2, 3t) - (2, 3) \right] \right \rVert
\]
\[
= \left \lVert \frac{1}{t} \left[ (t^2, 0) \right] \right \rVert
\]
\[
= \left \lVert (t, 0) \right \rVert = |t| < \delta = \epsilon
\]
**Why is \( Lu = (2, 3) \)?**
\[
L = \begin{bmatrix}
\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F79599c56-a340-49a0-b0ff-829b3947a798%2Fcd2ee7d6-0063-4cc9-ab9a-f3fe1b88cefc%2Fxtsl84h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Example:**
Let \( f: \mathbb{R}^2 \to \mathbb{R}^2 \) be defined as \( f(x, y) = \left( x^2 y, (3x + 2)y \right) \). Consider vectors \( u = (1, 0) \) and \( c = (1, 1) \).
**Claim:** \( Lu = (2, 3) \)
**Proof:**
Fix \(\epsilon > 0\). Choose \(\delta = \frac{\epsilon}{2}\). Assume \(0 < |t| < \delta\). Then:
\[
\frac{1}{t} \left[ f(c + tu) - f(c) - Lu \right] = \left \lVert \frac{1}{t} \left[ f(1+t, 1) - (1, 5) - (2, 3) \right] \right \rVert
\]
\[
= \left \lVert \frac{1}{t} \left[ ((1+t)^2, 3(1+t)+2) - (1, 5) \right] - (2, 3) \right \rVert
\]
\[
= \left \lVert \frac{1}{t} \left[ (1 + 2t + t^2, 3t + 5) - (1, 5) \right] - (2, 3) \right \rVert
\]
\[
= \left \lVert \frac{1}{t} \left[ (2t + t^2, 3t) - (2, 3) \right] \right \rVert
\]
\[
= \left \lVert \frac{1}{t} \left[ (t^2, 0) \right] \right \rVert
\]
\[
= \left \lVert (t, 0) \right \rVert = |t| < \delta = \epsilon
\]
**Why is \( Lu = (2, 3) \)?**
\[
L = \begin{bmatrix}
\frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\
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We have to find the value of and prove it using definition.
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